Question

The equation of the plane bisecting the angle between the planes $3x+4y=4$ and $6x-2y+3z+5=0$ that contains the origin, is

• A. $9x-38y+15z+43=0$
• B. $51x+18y+15z=3$
• C. $9x+2y+3z+1=0$
• D. $17x+9y+15z=26$

Answer 1

The correct option is B $51x+18y+15z=3$

Equation of given planes can be written as

$3x+4y=4,6x-2y+3z+5=0$

formula is

$\frac{a_{1}x+b_{1}y+c_{1}z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}$ = + or -$\frac{a_{2}x+b_{2}y+c_{2}z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$

by substituting the values in the given formula we will get

$\frac{3x+4y+0z+-41}{\sqrt{3^2+4^2+0^2}}$ = + or -$\frac{6x+-2y+3z+5}{\sqrt{6^2+(-2{)}^2+3^2}}$

$\Rightarrow$ $21x+28y-28=+or-30x-10y+160+25$

so when adding the above equation we will get $51x+18y+160z-3=0$

is the plane bisecting the angle containing the origin, and when subtracting we will get $9x-38y+160z+53=0$ is the other bisecting plane.

Hence the plane $51x+18y+160z-3=0or51x+18y+160z=3$ bisects the acute angle and therefore origin lies in the acute angle.